Problem: Let $A = (-3, 0),$ $B=(-2,1),$ $C=(2,1),$ and $D=(3,0).$ Suppose that point $P$ satisfies \[PA + PD = PB + PC = 8.\]Then the $y-$coordinate of $P,$ when simplified, can be expressed in the form $\frac{-a + b \sqrt{c}}{d},$ where $a,$ $b,$ $c,$ $d$ are positive integers.  Find $a + b + c + d.$
Since $PA + PD = 8,$ point $P$ must lie on the ellipse whose foci are $A$ and $D,$ and whose major axis has length $8.$ Since the distance between the foci is $3 - (-3) = 6,$ the minor axis has length $\sqrt{8^2 - 6^2} = 2\sqrt{7}.$ Then the semi-axes have lengths $4$ and $\sqrt{7},$ respectively, and the center of the ellipse is $(0,0),$ so the equation of this ellipse is \[\frac{x^2}{16} + \frac{y^2}{7} = 1.\]Similarly, since $PB+PC=8,$ point $P$ must lie on the ellipse whose foci are $B$ and $C,$ and whose major axis has length $8.$ Since the distance between the foci is $2-(-2) = 4,$ the minor axis has length $\sqrt{8^2-4^2} = 4\sqrt{3}.$ Then the semi-axes have lengths $4$ and $2\sqrt{3},$ respectively, and the center of the ellipse is $(0,1),$ so the equation of this ellipse is \[\frac{x^2}{16} + \frac{(y-1)^2}{12} = 1.\]Both ellipses are shown below. (Note that they intersect at two different points, but that they appear to have the same $y-$coordinate.) [asy] 
size(7cm);
pair A=(-3,0),B=(-2,1),C=(2,1),D=(3,0);
path ellipse1 = xscale(4)*yscale(sqrt(7))*unitcircle, ellipse2 = shift((0,1))*xscale(4)*yscale(sqrt(12))*unitcircle;
draw(ellipse1 ^^ ellipse2);
dot("$A$",A,S);
dot("$B$",B,S);
dot("$C$",C,S);
dot("$D$",D,S);
draw((-5,0)--(5,0),EndArrow); draw((0,-3.8)--(0,5.5),EndArrow);
label("$x$",(5,0),E); label("$y$",(0,5.5),N);
label("$\frac{x^2}{16}+\frac{y^2}{7}=1$",(3.2,5));
label("$\frac{x^2}{16}+\frac{(y-1)^2}{12}=1$",(3.4,-3));
pair [] p = intersectionpoints(ellipse1, ellipse2);
dot(p[0]^^p[1]);
[/asy]
Since $P$ lies on both ellipses, it must satisfy both equations, where $P=(x,y).$ We solve for $y.$ By comparing the two equations, we get \[\frac{y^2}{7} = \frac{(y-1)^2}{12}.\]Cross-multiplying and rearranging, we get the quadratic \[5y^2 + 14y - 7 = 0,\]and so by the quadratic formula, \[y=\frac{-14 \pm \sqrt{14^2 + 4 \cdot 5 \cdot 7}}{10} = \frac{-7 \pm 2\sqrt{21}}{5}.\]It remains to determine which value of $y$ is valid. Since $\sqrt{21} > 4,$ we have \[\frac{-7 - 2\sqrt{21}}{5} < \frac{-7 -2 \cdot 4}{5} = -3.\]But the smallest possible value of $y$ for a point on the ellipse $\frac{x^2}{16} + \frac{y^2}{7} = 1$ is $-\sqrt{7},$ which is greater than $-3.$ Therefore, we must choose the $+$ sign, and so \[y = \frac{-7 + 2\sqrt{21}}{5}.\]The final answer is $7 + 2 + 21 + 5 = \boxed{35}.$